Using MS Access and I have two tables, one is categories and the other is content.
My initial SQL statement, included below,takes a count of each content associated to a category and returns the count associated with each category.
So for each CATEGORY, I'm simply trying to return another count in which I count CONTENT that have a specific user level and are not deleted for each CATEGORY.
Below is what I am struggling with as I am not certain you can actually use COUNT like this.
COUNT(IIf([CONTENT.isDeleted]=0,1,0)) - COUNT(IIf([CONTENT.userLevel]=2)) AS userLevelCount
This is the full select statement with my addition but not working.
SELECT
CATEGORY.categoryId,
CATEGORY.categoryTitle,
CATEGORY.categoryDate,
CATEGORY.userLevel,
Last(CONTENT.contentDate) AS contentDate,
CATEGORY.isDeleted AS categoryDeleted,
COUNT(IIf([CONTENT.isDeleted]=0,1,0)) AS countTotal,
COUNT(IIf([CONTENT.isDeleted]=1,[CONTENT.contentID],Null)) AS countDeleted,
COUNT([CONTENT.categoryId]) - COUNT(IIf([CONTENT.isDeleted]=1,[CONTENT.contentID],Null))AS countDifference,
COUNT(IIf([CONTENT.isDeleted]=0,1,0)) - COUNT(IIf([CONTENT.userLevel]=2)) AS userLevelCount
FROM CATEGORY
LEFT JOIN CONTENT ON
CATEGORY.categoryId = CONTENT.categoryId
GROUP BY
CATEGORY.categoryId,
CATEGORY.categoryTitle,
CATEGORY.categoryDate,
CATEGORY.userLevel,
CATEGORY.isDeleted
HAVING (((CATEGORY.isDeleted)=0))
ORDER BY
CATEGORY.categoryTitle
you should be able to use the following
SUM(IIf([CONTENT.isDeleted]=0,1,0)) - COUNT(IIf([CONTENT.userLevel]=2,1,NULL)) AS userLevelCount
COUNT will not count NULL, but it will count zero. SUM will calculate the sum of all 1's - that's a second way of achieving the same.
IIF exists in the newer SQL versions
I believe I found the solution
Count(IIf([CONTENT.userLevel]=2,[CONTENT.contentID],Null)) AS countDifference2
This will return the count difference for CONTENT for each CATEGORY that isn't deleted and has a specific user level.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.